Why should we care about sufficient statistics? So I just learned the concept of sufficient statistics and I don't entirely understand why it's important.
I can think of two reasons but I'd like to get a confirmation for them:


*

*Is it because it reduces the dimension of the data (from let's say $R^n$ to $R$ in the case where $\overline{X}$ is a sufficient statistic) which could be computationally significant?

*It's a sought out property of statistic like unbiasedness and consistency (although I'm not sure why we'd care about sufficiency while I understand why we care about the former two). 

 A: Your reasons are both correct. Sufficiency is 'sought out'
because, along with other conditions (unbiasedness and completeness), it helps to identify
estimators that have the smallest variance. The intuitive idea
is that for purposes of estimating the parameter the sufficient
statistic contains all relevant information. (For other purposes,
such as testing goodness-of-fit to a particular distribution, all
of the original data may be required.)
Uniform. For example, if you are trying to estimate $\theta$ in
$Unif(0, \theta),$ you might consider three candidate
estimators: (a) double the sample mean, (b) double the
sample median, and (c) $(n+1)/n$ times the maximum
observation among $n$. All three are unbiased. Only (c)
is sufficient and it has the smallest variance of the three.
Indeed, (c) has the smallest possible variance among
unbiased estimators. 
Specifically, when $n = 5$ and $\theta = 10,$ the standard deviations
of these estimators are 1.69 for (c), compared with
2.58 for (a) and 3.78 for (b). So there is a considerable
difference in the variability of these estimates. (I mention
standard deviations instead of variances because SDs have
the same units as the mean.)
Poisson. As another example, if you are trying to estimate the mean $\lambda$,
of a Poisson distribution, the sample mean
and the sample variance are both unbiased
estimators of $\lambda$, but the mean is sufficient
and has the smaller variance.
Specifically, if $n = 5$ and $\lambda = 10,$ then
the sample mean has SD 1.42 and the sample variance has SD 7.22.
Note: Completeness, mentioned above, is a somewhat technical property and, in general,
difficult to prove. I suspect you will encounter it soon, but
now may not be the time to think about it.
